Range and Null Space
Range and Null Space
Definition : Let T : V → W T:V \rightarrow W T : V → W be a linear transformation. The set of all vectors y ∈ W y \in W y ∈ W such that y = T ( x ) y = T(x) y = T ( x ) for some x ∈ V x \in V x ∈ V is called the range of T T T and denoted by R ( T ) \mathcal{R}(T) R ( T ) .
Definition : Let T : V → W T:V \rightarrow W T : V → W be a linear transformation. The set of all vectors x ∈ V x \in V x ∈ V such that T ( x ) = 0 T(x) = 0 T ( x ) = 0 is called the null space of T T T and denoted by N ( T ) \mathcal{N}(T) N ( T ) .
Let T : V → W T:V \rightarrow W T : V → W be a linear transformation. Let d i m ( V ) = n dim(V) = n d im ( V ) = n and d i m ( W ) = m dim(W) = m d im ( W ) = m . Recall that,
R ( T ) : Range space of T , R ( T ) ⊆ W , d i m ( R ( T ) ) = d i m ( c o l ( T ) ) \mathcal{R}(T): \text{Range space of } T, \mathcal{R}(T) \subseteq W, dim(\mathcal{R}(T)) = dim(col(T)) R ( T ) : Range space of T , R ( T ) ⊆ W , d im ( R ( T )) = d im ( co l ( T ))
N ( T ) : Null space of T , N ( T ) ⊆ V , d i m ( N ( T ) ) = d i m ( n u l l ( T ) ) \mathcal{N}(T): \text{Null space of } T, \mathcal{N}(T) \subseteq V, dim(\mathcal{N}(T)) = dim(null(T)) N ( T ) : Null space of T , N ( T ) ⊆ V , d im ( N ( T )) = d im ( n u ll ( T ))
Theorem : d i m ( R ( T ) ) + d i m ( N ( T ) ) = d i m ( V ) dim(\mathcal{R}(T)) + dim(\mathcal{N}(T)) = dim(V) d im ( R ( T )) + d im ( N ( T )) = d im ( V )
Proof : Let B = { v 1 , v 2 , … , v k } B = \{v_1, v_2, \dots, v_k\} B = { v 1 ​ , v 2 ​ , … , v k ​ } be a basis for N ( T ) \mathcal{N}(T) N ( T ) . Extend B B B to a basis B ′ = { v 1 , v 2 , … , v k , v k + 1 , … , v n } B' = \{v_1, v_2, \dots, v_k, v_{k+1}, \dots, v_n\} B ′ = { v 1 ​ , v 2 ​ , … , v k ​ , v k + 1 ​ , … , v n ​ } for V V V . Then B ′ B' B ′ is a basis for V V V and B ′ ′ = { T ( v k + 1 ) , T ( v k + 2 ) , … , T ( v n ) } B'' = \{T(v_{k+1}), T(v_{k+2}), \dots, T(v_n)\} B ′′ = { T ( v k + 1 ​ ) , T ( v k + 2 ​ ) , … , T ( v n ​ )} is a basis for R ( T ) \mathcal{R}(T) R ( T ) . Hence d i m ( R ( T ) ) = n − k = d i m ( V ) − d i m ( N ( T ) ) dim(\mathcal{R}(T)) = n - k = dim(V) - dim(\mathcal{N}(T)) d im ( R ( T )) = n − k = d im ( V ) − d im ( N ( T )) â– \blacksquare â–
Definition : Let T : V → W T:V \rightarrow W T : V → W be a linear transformation. The rank of T T T is defined as r a n k ( T ) = d i m ( R ( T ) ) rank(T) = dim(\mathcal{R}(T)) r ank ( T ) = d im ( R ( T )) .
Remark : r a n k ( T ) rank(T) r ank ( T ) is equal to the number of linearly independent columns of T T T .
Remark : In general, a basis for R ( T ) \mathcal{R}(T) R ( T ) and N ( T ) \mathcal{N}(T) N ( T ) can be found by finding a basis for the column space and null space of the matrix representation of T T T with respect to some basis for V V V and W W W .
Eample : V = R 2 x 2 V = \mathbb{R}^{2x2} V = R 2 x 2 , W = V W = V W = V , Find a basis for R ( A ) \mathcal{R}(\mathcal{A}) R ( A )
B = { [ 1 0 0 0 ] , [ 0 1 0 0 ] , [ 0 0 1 0 ] , [ 0 0 0 1 ] } C = { [ 1 0 0 0 ] , [ 1 1 0 0 ] , [ 1 1 1 0 ] , [ 1 1 1 1 ] } A ( x ) = S x + x S ⊤ = [ 0 1 − 1 0 ] x + x [ 0 1 − 1 0 ] ⊤ A = [ 1 1 1 − 1 0 0 0 0 − 1 1 1 1 0 − 1 − 1 0 ] \begin{align*}
B &= \begin{Bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \end{Bmatrix} \\
C &= \begin{Bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \end{Bmatrix} \\
\mathcal{A}(x) &= Sx + x S^\top = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}x + x\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}^\top \\
A &= \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ -1 & 1 & 1 & 1 \\ 0 & -1 & -1 & 0 \end{bmatrix} \\
\end{align*} B C A ( x ) A ​ = { [ 1 0 ​ 0 0 ​ ] , [ 0 0 ​ 1 0 ​ ] , [ 0 1 ​ 0 0 ​ ] , [ 0 0 ​ 0 1 ​ ] ​ } = { [ 1 0 ​ 0 0 ​ ] , [ 1 0 ​ 1 0 ​ ] , [ 1 1 ​ 1 0 ​ ] , [ 1 1 ​ 1 1 ​ ] ​ } = S x + x S ⊤ = [ 0 − 1 ​ 1 0 ​ ] x + x [ 0 − 1 ​ 1 0 ​ ] ⊤ = ⎣ ⎡ ​ 1 0 − 1 0 ​ 1 0 1 − 1 ​ 1 0 1 − 1 ​ − 1 0 1 0 ​ ⎦ ⎤ ​ ​
Solution : Start with finding a basis for R ( A ) \mathcal{R}(\mathcal{A}) R ( A )
R ( A ) : = { w ∈ W  ∣   ∃ v ∈ V ,  A v = w } R ( A ) : = { y ∈ R 4  ∣   ∃ x ∈ R 4 ,  A x = y } \begin{align*}
\mathcal{R}(\mathcal{A}) &:= \{ w \in W \ | \ \ \exist v \in V , \ \mathcal{A}v = w \} \\
\mathcal{R}(A) &:= \{ y \in \mathbb{R}^4 \ | \ \ \exist x \in \mathbb{R}^4 , \ Ax = y \} \\
\end{align*} R ( A ) R ( A ) ​ := { w ∈ W  ∣   ∃ v ∈ V ,  A v = w } := { y ∈ R 4  ∣   ∃ x ∈ R 4 ,  A x = y } ​
Start with finding a basis for R ( A ) \mathcal{R}(A) R ( A )
Convert the vectors in the basis into matrices using the basis C C C .
Finding a basis for R ( A ) \mathcal{R}(A) R ( A ) is equivalent to finding a basis for the column space of A A A . The basis for the column space of A A A can be found by reducing A A A to independent columns by elementary column operations and then taking the non-zero columns of the reduced matrix.
A = [ 1 1 1 − 1 0 0 0 0 − 1 1 1 1 0 − 1 − 1 0 ] ∼ [ 1 0 0 0 0 0 0 0 − 1 2 0 0 0 − 1 0 0 ] \begin{align*}
A &= \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ -1 & 1 & 1 & 1 \\ 0 & -1 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix} \\
\end{align*} A ​ = ⎣ ⎡ ​ 1 0 − 1 0 ​ 1 0 1 − 1 ​ 1 0 1 − 1 ​ − 1 0 1 0 ​ ⎦ ⎤ ​ ∼ ⎣ ⎡ ​ 1 0 − 1 0 ​ 0 0 2 − 1 ​ 0 0 0 0 ​ 0 0 0 0 ​ ⎦ ⎤ ​ ​
basis for R ( A ) = { [ 1 0 − 1 0 ] , [ 0 0 2 − 1 ] } \begin{align*}
\text{basis for } \mathcal{R}(A) &= \begin{Bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 2 \\ -1 \end{bmatrix}\end{Bmatrix} \\
\end{align*} basis for R ( A ) ​ = ⎩ ⎨ ⎧ ​ ⎣ ⎡ ​ 1 0 − 1 0 ​ ⎦ ⎤ ​ , ⎣ ⎡ ​ 0 0 2 − 1 ​ ⎦ ⎤ ​ ​ ⎠⎬ ⎫ ​ ​
[ 1 0 − 1 0 ] = [ w 1 ] C    ⟹    w 1 = 1 c 1 + 0 c 2 + ( − 1 ) c 3 + 0 c 4 = c 1 − c 3 = [ 1 0 0 0 ] − [ 1 1 1 0 ] = [ 0 − 1 − 1 0 ] \begin{align*}
\begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix} = [w_1]_C \implies w_1 &= 1c_1 + 0c_2 + (-1)c_3 + 0c_4 = c_1 - c_3 \\
&= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \\
\end{align*} ⎣ ⎡ ​ 1 0 − 1 0 ​ ⎦ ⎤ ​ = [ w 1 ​ ] C ​ ⟹ w 1 ​ ​ = 1 c 1 ​ + 0 c 2 ​ + ( − 1 ) c 3 ​ + 0 c 4 ​ = c 1 ​ − c 3 ​ = [ 1 0 ​ 0 0 ​ ] − [ 1 1 ​ 1 0 ​ ] = [ 0 − 1 ​ − 1 0 ​ ] ​
[ 0 0 2 − 1 ] = [ w 2 ] C    ⟹    w 2 = 0 c 1 + 0 c 2 + 2 c 3 + ( − 1 ) c 4 = 2 c 3 − c 4 = [ 2 2 2 0 ] − [ 1 1 1 1 ] = [ 1 1 1 − 1 ] \begin{align*}
\begin{bmatrix} 0 \\ 0 \\ 2 \\ -1 \end{bmatrix} = [w_2]_C \implies w_2 &= 0c_1 + 0c_2 + 2c_3 + (-1)c_4 = 2c_3 - c_4 \\
&= \begin{bmatrix} 2 & 2 \\ 2 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \\
\end{align*} ⎣ ⎡ ​ 0 0 2 − 1 ​ ⎦ ⎤ ​ = [ w 2 ​ ] C ​ ⟹ w 2 ​ ​ = 0 c 1 ​ + 0 c 2 ​ + 2 c 3 ​ + ( − 1 ) c 4 ​ = 2 c 3 ​ − c 4 ​ = [ 2 2 ​ 2 0 ​ ] − [ 1 1 ​ 1 1 ​ ] = [ 1 1 ​ 1 − 1 ​ ] ​
basis for R ( A ) = { [ 0 − 1 − 1 0 ] , [ 1 1 1 − 1 ] } d i m ( R ( A ) ) = 2 = rank ( A ) \begin{align*}
\text{basis for } \mathcal{R}(\mathcal{A}) &= \begin{Bmatrix}\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\end{Bmatrix} \\
dim(\mathcal{R}(A)) &= 2 = \text{rank}(A) \\
\end{align*} basis for R ( A ) d im ( R ( A )) ​ = { [ 0 − 1 ​ − 1 0 ​ ] , [ 1 1 ​ 1 − 1 ​ ] ​ } = 2 = rank ( A ) ​
A A A is a 4 x 4 4x4 4 x 4 matrix, d i m ( R ( A ) ) = 4 − d i m ( N ( A ) ) dim(\mathcal{R}(A)) = 4 - dim(\mathcal{N}(A)) d im ( R ( A )) = 4 − d im ( N ( A ))
d i m ( N ( A ) ) = 2 dim(\mathcal{N}(A)) = 2 d im ( N ( A )) = 2 since r a n k ( A ) = 2 rank(A) = 2 r ank ( A ) = 2
d i m ( R ( A ) ) = 4 − 2 = 2 dim(\mathcal{R}(A)) = 4 - 2 = 2 d im ( R ( A )) = 4 − 2 = 2
d i m ( R ( A ) ) = 2 dim(\mathcal{R}(\mathcal{A})) = 2 d im ( R ( A )) = 2
Now, we need to find a basis for N ( A ) \mathcal{N}(\mathcal{A}) N ( A ) .
N ( A ) : = { v ∈ V  ∣   A v = 0 } N ( A ) : = { x ∈ R 4  ∣   A x = 0 W } \begin{align*}
\mathcal{N}(\mathcal{A}) &:= \{ v \in V \ | \ \ \mathcal{A}v = 0 \} \\
\mathcal{N}(A) &:= \{ x \in \mathbb{R}^4 \ | \ \ Ax = 0_W \} \\
\end{align*} N ( A ) N ( A ) ​ := { v ∈ V  ∣   A v = 0 } := { x ∈ R 4  ∣   A x = 0 W ​ } ​
Start with finding a basis for N ( A ) \mathcal{N}(A) N ( A )
Convert the vectors in the basis into matrices using the basis B B B .
Finding a basis for N ( A ) \mathcal{N}(A) N ( A ) is equivalent to finding a basis for the null space of A A A . The basis for the null space of A A A can be found by reducing A A A by elementary row operations and then equating the A ˉ x = 0 \bar Ax =0 A ˉ x = 0 .
A = [ 1 1 1 − 1 0 0 0 0 − 1 1 1 1 0 − 1 − 1 0 ] ∼ [ 1 0 0 − 1 0 1 1 0 0 0 0 0 0 0 0 0 ] = A ˉ \begin{align*}
A &= \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ -1 & 1 & 1 & 1 \\ 0 & -1 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \bar A \\
\end{align*} A ​ = ⎣ ⎡ ​ 1 0 − 1 0 ​ 1 0 1 − 1 ​ 1 0 1 − 1 ​ − 1 0 1 0 ​ ⎦ ⎤ ​ ∼ ⎣ ⎡ ​ 1 0 0 0 ​ 0 1 0 0 ​ 0 1 0 0 ​ − 1 0 0 0 ​ ⎦ ⎤ ​ = A ˉ ​
A ˉ x = 0 [ 1 0 0 − 1 0 1 1 0 0 0 0 0 0 0 0 0 ] [ x 1 x 2 x 3 x 4 ] = [ 0 0 0 0 ] x 1 − x 4 = 0    ⟹    x 1 = x 4 x 2 + x 3 = 0    ⟹    x 2 = − x 3 \begin{align*}
\bar Ax = 0 \\
\begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \\
x_1 - x_4 = 0 \implies x_1 &= x_4 \\
x_2 + x_3 = 0 \implies x_2 &= -x_3 \\
\end{align*} A ˉ x = 0 ⎣ ⎡ ​ 1 0 0 0 ​ 0 1 0 0 ​ 0 1 0 0 ​ − 1 0 0 0 ​ ⎦ ⎤ ​ ⎣ ⎡ ​ x 1 ​ x 2 ​ x 3 ​ x 4 ​ ​ ⎦ ⎤ ​ x 1 ​ − x 4 ​ = 0 ⟹ x 1 ​ x 2 ​ + x 3 ​ = 0 ⟹ x 2 ​ ​ = ⎣ ⎡ ​ 0 0 0 0 ​ ⎦ ⎤ ​ = x 4 ​ = − x 3 ​ ​
[ x 1 x 2 x 3 x 4 ] = [ x 1 x 2 − x 2 x 1 ] = x 1 [ 1 0 0 1 ] + x 2 [ 0 1 − 1 0 ] \begin{align*}
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} &= \begin{bmatrix} x_1 \\ x_2 \\ -x_2 \\ x_1 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix} \\
\end{align*} ⎣ ⎡ ​ x 1 ​ x 2 ​ x 3 ​ x 4 ​ ​ ⎦ ⎤ ​ ​ = ⎣ ⎡ ​ x 1 ​ x 2 ​ − x 2 ​ x 1 ​ ​ ⎦ ⎤ ​ = x 1 ​ ⎣ ⎡ ​ 1 0 0 1 ​ ⎦ ⎤ ​ + x 2 ​ ⎣ ⎡ ​ 0 1 − 1 0 ​ ⎦ ⎤ ​ ​
basis for N ( A ) = { [ 1 0 0 1 ] , [ 0 1 − 1 0 ] } \begin{align*}
\text{basis for } \mathcal{N}(A) &= \begin{Bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix}\end{Bmatrix} \\
\end{align*} basis for N ( A ) ​ = ⎩ ⎨ ⎧ ​ ⎣ ⎡ ​ 1 0 0 1 ​ ⎦ ⎤ ​ , ⎣ ⎡ ​ 0 1 − 1 0 ​ ⎦ ⎤ ​ ​ ⎠⎬ ⎫ ​ ​
[ 1 0 0 1 ] = [ v 1 ] B    ⟹    v 1 = 1 b 1 + 0 b 2 + 0 b 3 + 1 b 4 = b 1 + b 4 = [ 1 0 0 0 ] + [ 0 0 0 1 ] = [ 1 0 0 1 ] \begin{align*}
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} = [v_1]_B \implies v_1 &= 1b_1 + 0b_2 + 0b_3 + 1b_4 = b_1 + b_4 \\
&= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\
\end{align*} ⎣ ⎡ ​ 1 0 0 1 ​ ⎦ ⎤ ​ = [ v 1 ​ ] B ​ ⟹ v 1 ​ ​ = 1 b 1 ​ + 0 b 2 ​ + 0 b 3 ​ + 1 b 4 ​ = b 1 ​ + b 4 ​ = [ 1 0 ​ 0 0 ​ ] + [ 0 0 ​ 0 1 ​ ] = [ 1 0 ​ 0 1 ​ ] ​
[ 0 1 − 1 0 ] = [ v 2 ] B    ⟹    v 2 = 0 b 1 + 1 b 2 + ( − 1 ) b 3 + 0 b 4 = b 2 − b 3 = [ 0 1 0 0 ] − [ 0 0 1 0 ] = [ 0 1 − 1 0 ] \begin{align*}
\begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix} = [v_2]_B \implies v_2 &= 0b_1 + 1b_2 + (-1)b_3 + 0b_4 = b_2 - b_3 \\
&= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \\
\end{align*} ⎣ ⎡ ​ 0 1 − 1 0 ​ ⎦ ⎤ ​ = [ v 2 ​ ] B ​ ⟹ v 2 ​ ​ = 0 b 1 ​ + 1 b 2 ​ + ( − 1 ) b 3 ​ + 0 b 4 ​ = b 2 ​ − b 3 ​ = [ 0 0 ​ 1 0 ​ ] − [ 0 1 ​ 0 0 ​ ] = [ 0 − 1 ​ 1 0 ​ ] ​
basis for N ( A ) = { [ 1 0 0 1 ] , [ 0 1 − 1 0 ] } d i m ( N ( A ) ) = 2 = nullity ( A ) \begin{align*}
\text{basis for } \mathcal{N}(\mathcal{A}) &= \begin{Bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\end{Bmatrix} \\
dim(\mathcal{N}(A)) &= 2 = \text{nullity}(A) \\
\end{align*} basis for N ( A ) d im ( N ( A )) ​ = { [ 1 0 ​ 0 1 ​ ] , [ 0 − 1 ​ 1 0 ​ ] ​ } = 2 = nullity ( A ) ​
Definition : Let T : V → W T:V \rightarrow W T : V → W be a linear transformation. The nullity of T T T is defined as n u l l i t y ( T ) = d i m ( N ( T ) ) nullity(T) = dim(\mathcal{N}(T)) n u ll i t y ( T ) = d im ( N ( T )) .
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